What is the answer to log_{2}(2^{x+1}-32sqrt(2))=x ?

The answer to log_{2}(2^{x+1}-32sqrt(2))=x is x= 11/2

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log_{2}(2^{x+1}-32sqrt(2))=x

lo g_{2} (2^{x + 1} - 322) = x

x = \frac{11}{2}

Decimal Notation

x = 5.5

Solution steps

lo g_{2} (2^{x + 1} - 322) = x

f (x) = g (x),

then

a^{f (x)} = a^{g (x)}

2^{l o g_{2} (2^{x + 1} - 322)} = 2^{x}

Simplify

2^{l o g_{2} (2^{x + 1} - 322)} : 2^{x + 1} - 322

2^{x + 1} - 322 = 2^{x}

Solve

2^{x + 1} - 322 = 2^{x} : x = \frac{11}{2}

x = \frac{11}{2}

Verify Solutions:

x = \frac{11}{2}

True

The solution is

x = \frac{11}{2}

ln (y - 1) = 1 + ln (3 y + 2)

lo g_{3} (x + 1) = 3

lo g_{3} (2 x - 5) = 2

lo g_{10} (5) + lo g_{10} (x) = 2

ln (x) + ln (x + 1) = 0

What is the answer to log_{2}(2^{x+1}-32sqrt(2))=x ?
The answer to log_{2}(2^{x+1}-32sqrt(2))=x is x= 11/2
Solve for x: log_{2}(2^{x+1}-32sqrt(2))=x
The solution is x= 11/2